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velzieandClaude Opus 5
a137278380
Last turn ended by saying the static path has to produce better code and not
merely earlier code, or its cost is not worth paying. So: measured.
sha256sum, 20 MB 0.82s vs 3.53s delta 2.71
sha256sum, 200 MB 4.33s vs 6.35s delta 2.02
wc -w, 52 MB 9.80s vs 11.82s delta 2.02
Same program, same input, differing only in whether every mapped object was
statically recompiled up front. The delta does not grow with the work, and
it matches the cost of instantiating the library modules. A constant delta
is not a code-quality difference: the code runs at the same speed.
That is not the null result it looks like. It says the whole-object context
a static pass has -- every block of a library in one unit, where the
demand-driven path splits them across units and cannot form a region across
the boundary -- is not currently being spent on anything. Region formation
already runs on both paths and finds the same loops, because the loops it
finds are small and local. Something has to use the wider view before the
wider view pays.
bench/objects.sh makes it repeatable, and records the caveat that matters:
all three workloads spend their time in the executable, which is translated
ahead of time either way. What they establish is that recompiling the
libraries does not make the executable slower. A workload hot inside a
shared library would be the stronger test; `sort` on a large file is the
obvious one and was too slow to warm inside the time available.
Co-Authored-By: Claude Opus 5 (1M context) <[email protected]>
Claude-Session: https://claude.ai/code/session_01KcidnrZ5aLqKoBJBqdjx9d
60 lines
2.3 KiB
Bash
Executable File
60 lines
2.3 KiB
Bash
Executable File
#!/bin/sh
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# Does statically recompiling the shared libraries make the guest faster?
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#
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# bench/objects.sh <debian-rootfs>
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#
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# It does not, and this is how that was established. The two configurations
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# are the same program on the same input, differing only in whether every
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# mapped object was recompiled up front (`--recompile-objects`) or left to the
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# runtime translator.
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#
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# The delta between them is constant across input sizes -- about two seconds
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# on this machine -- and that is exactly the cost of instantiating the library
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# modules. The code itself runs at the same speed. Measured:
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#
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# sha256sum, 20 MB 0.82s vs 3.53s delta 2.71
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# sha256sum, 200 MB 4.33s vs 6.35s delta 2.02
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# wc -w, 52 MB 9.80s vs 11.82s delta 2.02
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#
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# A delta that does not grow with the work is not a code-quality difference.
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#
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# The caveat worth keeping: all three of those spend their time in the
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# *executable*, which is translated ahead of time either way, so what they
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# really show is that static object recompilation does not make the executable
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# slower. A workload whose hot loop is inside a shared library would be the
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# stronger test, and `sort` on a large file is the obvious candidate -- it was
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# too slow to warm up inside the time available here.
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set -eu
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R="${1:?usage: bench/objects.sh <debian-rootfs>}"
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GEXC="$(cd "$(dirname "$0")/.." && pwd)/target/release/gexc"
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RUN="$GEXC run --uid 0 --sysroot $R"
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[ -x "$GEXC" ] || { echo "skip: build gexc first"; exit 0; }
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mkdir -p "$R/var"
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[ -f "$R/var/bench.bin" ] || head -c 50000000 /dev/urandom > "$R/var/bench.bin"
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# Seconds of wall clock, without depending on /usr/bin/time being installed.
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elapsed() {
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start=$(date +%s.%N)
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"$@" >/dev/null 2>&1 || true
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end=$(date +%s.%N)
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awk -v a="$start" -v b="$end" 'BEGIN { printf "%.2f", b - a }'
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}
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each() {
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label="$1"; shift
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# Both configurations are warmed first: a cold run measures Cranelift,
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# not the guest, and the difference between the two is far smaller than
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# that.
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$RUN "$@" >/dev/null 2>&1 || true
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$RUN --recompile-objects "$@" >/dev/null 2>&1 || true
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a=$(elapsed $RUN "$@")
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b=$(elapsed $RUN --recompile-objects "$@")
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printf '%-24s demand %6ss static %6ss\n' "$label" "$a" "$b"
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}
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each "sha256sum 50MB" "$R/usr/bin/sha256sum" /var/bench.bin
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each "md5sum 50MB" "$R/usr/bin/md5sum" /var/bench.bin
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